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Do now!. DEFINITIONS TEST!! You have 12 minutes!. Topic 4 Oscillations and Waves. Aims. Remember the terms displacement, amplitude, frequency, period and phase difference. Define simple harmonic motion (a = - ω 2 x) Solve problems using a = - ω 2 x

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  1. Do now! DEFINITIONS TEST!! You have 12 minutes!

  2. Topic 4 Oscillations and Waves

  3. Aims • Remember the terms displacement, amplitude, frequency, period and phase difference. • Define simple harmonic motion (a = -ω2x) • Solve problems using a = -ω2x • Apply the equations x = x0cosωt, x = x0sinωt, v = v0sinωt, v = v0cosωt, and v = ±ω√(x02 – x2)

  4. Displacement - x The distance and direction from the equilibrium position. = displacement

  5. Amplitude - A The maximum displacement from the equilibrium position. amplitude

  6. Period - T The time taken (in seconds) for one complete oscillation. It is also the time taken for a complete wave to pass a given point. One complete wave

  7. Frequency - f The number of oscillations in one second. Measured in Hertz. 50 Hz = 50 vibrations/waves/oscillations in one second.

  8. Period and frequency Period and frequency are reciprocals of each other f = 1/T T = 1/f

  9. Phase difference • is the time difference or phase angle by which one wave/oscillation leads or lags another. 180° or π radians

  10. Phase difference • is the time difference or phase angle by which one wave/oscillation leads or lags another. 90° or π/2 radians

  11. Simple harmonic motion (SHM) • periodic motion in which the restoring force is proportional and in the opposite direction to the displacement

  12. Hooke’s law What can you remember?

  13. Simple harmonic motion (SHM) • periodic motion in which the restoring force is in the opposite durection and proportional to the displacement F = -kx

  14. displacement Time Graph of motion A graph of the motion will have this form

  15. displacement Time Graph of motion A graph of the motion will have this form Amplitude x0 Period T

  16. Graph of motion Notice the similarity with a sine curve 2π radians angle 3π/2 2π π π/2

  17. Graph of motion Notice the similarity with a sine curve Amplitude x0 x = x0sinθ 2π radians angle 3π/2 2π π π/2

  18. displacement Time Graph of motion Amplitude x0 Period T

  19. displacement Time Graph of motion Amplitude x0 Period T x = x0sinωt where ω = 2π/T = 2πf = (angular frequency in rad.s-1)

  20. displacement Time When x = 0 at t = 0 Amplitude x0 Period T x = x0sinωt where ω = 2π/T = 2πf = (angular frequency in rad.s-1)

  21. When x = x0 at t = 0 Amplitude x0 Period T x = x0cosωt displacement Time where ω = 2π/T = 2πf = (angular frequency in rad.s-1)

  22. displacement Time When x = 0 at t = 0 x = x0sinωt v = v0cosωt Amplitude x0 Period T where ω = 2π/T = 2πf = (angular frequency in rad.s-1)

  23. When x = x0 at t = 0 x = x0cosωt v = -v0sinωt Amplitude x0 Period T displacement Time where ω = 2π/T = 2πf = (angular frequency in rad.s-1)

  24. To summarise! • When x = 0 at t = 0 x = x0sinωtand v = v0cosωt • When x = x0 at t = 0 x = x0cosωt andv = -v0sinωt It can also be shown thatv = ±ω√(x02 – x2) and a = -ω2x where ω = 2π/T = 2πf = (angular frequency in rad.s-1)

  25. Maximum velocity?

  26. Maximum velocity? • When x = 0 • At this point the acceleration is zero (no resultant force at the equilibrium position).

  27. Maximum acceleration?

  28. Maximum acceleration? • When x = +/– x0 • Here the velocity is zero

  29. Oscillating spring We know that F = -kx and that for SHM, a = -ω2x (so F = -mω2x) So -kx = -mω2x k = mω2 ω = √(k/m) Remembering that ω = 2π/T T = 2π√(m/k)

  30. Let’s do a simple practical! T = 2π√(m/k)

  31. We need to try some examples!

  32. Example 1 Find the acceleration of a system oscillating with SHM where ω = 2.5 rad s-1 and x = 0.5 m to 2.s.f.

  33. Example 1 Find the acceleration of a system oscillating with SHM where ω = 2.5 rad s-1 and x = 0.5 m to 2.s.f. Using a = -ω2x a = -(2.5)20.5 a = -3.13 m.s-2

  34. Example 2 Find the displacement at a point to 2.s.f. of a system of frequency 4 Hz when its acceleration is -8 ms-2.

  35. Example 2 Find the displacement at a point to 2.s.f. of a system of frequency 4 Hz when its acceleration is -8 ms-2. ω = 2πf = 2π x 4 = 8π a = -ω2x x = -a/ω2 = 8/(8π)2 = 1/8π2 = 0.013 m

  36. Example 3 For a SHM system of x = 0 at t = 0, find x when ω = 5.0 rad s-1, xo = 0.5 m and t = 1.0 s. What is the maximum acceleration of this system to 3.s.f? What is the maximum velocity of this system?

  37. Example 3 For a SHM system of x = 0 at t = 0, find x when ω = 5.0 rad s-1, xo = 0.5 m and t = 1.0 s. What is the maximum acceleration of this system to 3.s.f? What is the maximum velocity of this system? x = xosinωt (when x = 0 at t = 0) x = 0.5sin(5.0 x 1.0) = 0.5sin5 = -0.479 m

  38. Example 3 For a SHM system of x = 0 at t = 0, find x when ω = 5.0 rad s-1, xo = 0.5 m and t = 1.0 s. What is the maximum acceleration of this system to 3.s.f? What is the maximum velocity of this system? a = -ω2x Maximum acceleration when x = ±xo amax = -ω2xo = -(5)2 x 0.5 = -12.5 m.s-2

  39. Example 3 For a SHM system of x = 0 at t = 0, find x when ω = 5.0 rad s-1, xo = 0.5 m and t = 1.0 s. What is the maximum acceleration of this system to 3.s.f? What is the maximum velocity of this system? v = ±ω√(x02 – x2) Maximum velocity when x = 0 vmax = ± ω√(x02 – x2) = ±5.0√(0.5)2 = ±2.5 m.s-1

  40. Let’s try some questions! Finish for homework. Due Thursday 26th February (two days before Mr Porter’s virtual birthday)

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